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Electronic Devices - Special Purpose Junction Diodes (Zener, LED, Photodiode)

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A Zener Diode is a heavily doped pnp-n junction diode designed to operate specifically in the reverse breakdown region. Because of heavy doping, the depletion layer is very thin (approx106 m\\approx 10^{-6} \text{ m}), leading to a high electric field even at low voltages.

The Zener Breakdown occurs when the high electric field across the junction pulls electrons out of the covalent bonds, creating a large number of charge carriers. It is used for Voltage Regulation because the voltage across the diode remains constant (VZV_Z) even if the current through it changes significantly.

A Light Emitting Diode (LED) is a forward-biased pnp-n junction that emits spontaneous radiation. When the diode is forward biased, electrons from the nn-side and holes from the pp-side move toward the junction and recombine. The energy released during recombination is emitted as photons if the energy gap EgE_g is in the visible or infrared range.

For an LED, the wavelength of light emitted is given by λ=hcEg\lambda = \frac{hc}{E_g}. The color of light depends on the semiconductor material used (e.g., GaAsPGaAsP for red/yellow light).

A Photodiode is a special purpose pnp-n junction diode operated in reverse bias. It is used to detect optical signals. When light of frequency ν\nu such that hν>Egh\nu > E_g falls on the junction, it creates electron-hole pairs near the depletion region, increasing the reverse saturation current.

In a photodiode, the reverse current is directly proportional to the intensity of incident light. It is operated in reverse bias because the fractional change in minority carrier current is much more easily measurable than the fractional change in majority carrier current.

📐Formulae

Vs=VinVZV_s = V_{in} - V_Z

Is=VinVZRsI_s = \frac{V_{in} - V_Z}{R_s}

Is=IZ+ILI_s = I_Z + I_L

Eg=hν=hcλE_g = h\nu = \frac{hc}{\lambda}

λ(in nm)1240Eg(in eV)\lambda (\text{in nm}) \approx \frac{1240}{E_g (\text{in eV})}

💡Examples

Problem 1:

A Zener diode with a breakdown voltage VZ=6 VV_Z = 6 \text{ V} is used as a voltage regulator in a circuit. If the input voltage VinV_{in} varies between 10 V10 \text{ V} and 16 V16 \text{ V} and the series resistance RsR_s is 2 kΩ2 \text{ k}\Omega, calculate the maximum current through the Zener diode, assuming the load resistance RLR_L is very high (open circuit).

Solution:

Given VZ=6 VV_Z = 6 \text{ V}, Rs=2 kΩ=2000 ΩR_s = 2 \text{ k}\Omega = 2000 \text{ }\Omega. The maximum current occurs when the input voltage is at its maximum, Vin=16 VV_{in} = 16 \text{ V}. Since RLR_L is high, IL0I_L \approx 0, thus IZ=IsI_Z = I_s. Using the formula Is=VinVZRsI_s = \frac{V_{in} - V_Z}{R_s}, we get: IZ=1662000=102000=5×103 A=5 mAI_Z = \frac{16 - 6}{2000} = \frac{10}{2000} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}.

Explanation:

In a Zener regulator, the voltage across the Zener remains fixed at VZV_Z. The excess voltage VinVZV_{in} - V_Z drops across the series resistor RsR_s. The current is maximum when the source voltage is at its peak.

Problem 2:

A semiconductor has a bandgap of Eg=2.8 eVE_g = 2.8 \text{ eV}. Is it suitable for fabricating an LED that emits visible light?

Solution:

The wavelength of light emitted is λ=hcEg\lambda = \frac{hc}{E_g}. Using the approximation λ1240Eg nm\lambda \approx \frac{1240}{E_g} \text{ nm}, we get λ=12402.8442.8 nm\lambda = \frac{1240}{2.8} \approx 442.8 \text{ nm}.

Explanation:

The visible spectrum ranges from approximately 400 nm400 \text{ nm} to 700 nm700 \text{ nm}. Since 442.8 nm442.8 \text{ nm} falls within this range (corresponding to violet/blue light), the material is suitable for a visible light LED.

Special Purpose Junction Diodes (Zener, LED, Photodiode) Revision - Class 12 Physics ICSE